Jump to content

calculus BC help


encipher

Recommended Posts

I'm going to assume that you mean the sequence:

 

[math](a_n)_{n=1}^{\infty} = \frac{1}{n \log^2 n}[/math]

 

instead of the continuous function. But the same principle should apply, I think. You should notice that, for [imath]n \geq 2[/imath], [imath]0 \leq \frac{1}{n} \leq 1[/imath]. So,

 

[math]0 \leq \frac{1}{n \log^2 n} \leq \frac{1}{\log^2 n}[/math]

 

The rest should be obvious.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.